Periodic time of a satellite revolving above Earth's surface at a height equal to R, radius of Earth, is : [g is acceleration due to gravity at Earth's surface]

Options

A. `2 pi rsqrt((2R)/(g))`
B. `4 sqrt(2)pi sqrt((R)/(g))`
C. `2 pi sqrt((R)/(g))`
D. `8 pi sqrt((R)/(g))`

Explanation

To find the periodic time (T) of a satellite revolving above the Earth's surface at a height equal to the radius of the Earth (R), we can follow these steps:

### Step 1: Understand the distance of the satellite from the center of the Earth
The satellite is at a height equal to the radius of the Earth (R). Therefore, the total distance (d) from the center of the Earth to the satellite is:
\[ d = R + R = 2R \]

### Step 2: Write the expression for gravitational force
The gravitational force (F) acting on the satellite can be expressed using Newton's law of gravitation:
\[ F = \frac{G M m}{(2R)^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite.

### Step 3: Write the expression for centripetal force
The centripetal force required to keep the satellite in circular motion is given by:
\[ F_c = \frac{m v^2}{r} \]
where \( v \) is the orbital speed of the satellite and \( r \) is the radius of the orbit (which is \( 2R \)).

### Step 4: Set gravitational force equal to centripetal force
Since the gravitational force provides the necessary centripetal force for the satellite, we can set them equal:
\[ \frac{G M m}{(2R)^2} = \frac{m v^2}{2R} \]

### Step 5: Simplify the equation
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{4R^2} = \frac{v^2}{2R} \]
Now, multiply both sides by \( 2R \):
\[ \frac{2G M}{4R} = v^2 \]
This simplifies to:
\[ v^2 = \frac{G M}{2R} \]

### Step 6: Relate orbital speed to the period
The orbital speed \( v \) can also be expressed in terms of the period \( T \):
\[ v = \frac{2\pi (2R)}{T} \]
Substituting this into the equation gives:
\[ \left(\frac{2\pi (2R)}{T}\right)^2 = \frac{G M}{2R} \]

### Step 7: Solve for T
Squaring both sides:
\[ \frac{4\pi^2 (4R^2)}{T^2} = \frac{G M}{2R} \]
Cross-multiplying gives:
\[ 4\pi^2 (4R^2) = \frac{G M T^2}{2R} \]
Rearranging to solve for \( T^2 \):
\[ T^2 = \frac{8\pi^2 R^3}{G M} \]

### Step 8: Substitute \( g \) in terms of \( G \) and \( M \)
We know that the acceleration due to gravity at the Earth's surface is given by:
\[ g = \frac{G M}{R^2} \]
Thus, we can express \( G M \) as:
\[ G M = g R^2 \]

### Step 9: Substitute back into the equation for T
Substituting \( G M \) into the equation for \( T^2 \):
\[ T^2 = \frac{8\pi^2 R^3}{g R^2} \]
This simplifies to:
\[ T^2 = \frac{8\pi^2 R}{g} \]

### Step 10: Take the square root to find T
Taking the square root gives:
\[ T = 2\pi \sqrt{\frac{8R}{g}} \]
This can be simplified further:
\[ T = 4\sqrt{2}\pi \sqrt{\frac{R}{g}} \]

### Final Answer
The periodic time of the satellite is:
\[ T = 4\sqrt{2}\pi \sqrt{\frac{R}{g}} \]